Either the idea of the question flew completely over my head or at its core this is really simple arithmetic veiled with complicated wording.
A is 50 because there are 100 percentiles and 50 integers between 151 and 200 inclusive => every integer has exactly two percentiles. And by this logic B should be 2?
Correct me if I'm wrong.

The minimum number of integers that include more than 1 percentile group, which means that whats the lowest number of a scores that can have more than 1 percentile group. If all 400 people get the same score, then that's one integer value that has all the percentiles.

The minimum number of percentile groups that correspond to a score of 200, well... what if no one gets a 200? then that's no integer values at all.

thus choice A =1
choice B = 0

A is greater

Hey Carcass, I need your help because this seems a hylogliphics.

Solution?

Hi All,

Can someone please help with this question? I have few questions regarding the percentile concept, in general.

consider the data set :

A = {151,151,........151} --->400 times

What will be the percentile for score 151 and how many percentile groups will be there? If I have one more data set:

B = {151,151,.....151(399 times),200}

how many percentile groups will be there and what will be the percentile for 200 score?

Regards,
arorni

Can someone please reply for the question asked above?

Regards,
arorni

Carcass GreenlightTestPrep rx10 KarunMendiratta Ks1859

Please shed some light in this matter.

Can someone please help with the question asked above?

Regards,
arorni

There are 50 integers corresponding to 100 percentile. Even in the case of uniform distribution each integer would have 2 percentile groups.

Let us say everybody gets 164.The integer now corresponds to all percentiles (since there is no distribution). Hence minimum number of integers with more than 1 percentile group is 1.

Now in the same example no one scored a 200. Hence it would have no percentile groups attached to it. So its minimum possible number of percentile groups is 0.

Hence Quantity A is greater.


PS: The question was not clear but you can get an idea through this explanation

Hope this helps

There are 400 test scores distributed among 50 possible outcomes
(integers between 151 and 200, inclusive, which number 200 – 151 + 1 = 50 integers).
There is an average of 400 ÷ 50 = 8 scores per integer outcome, and there are 400 ÷ 100 = 4 scores in each percentile.
So, if all the scores were
completely evenly distributed with exactly 8 scores per integer, there would
be two percentile groups per integer outcome (0th and 1st percentiles at 151,
2nd and 3rd percentiles at 152, etc.). In that case, all 50 integers from 151 to
200 would correspond to more than one percentile group.

Reduce the number of integers corresponding to more than one percentile
group by bunching up the scores. Imagine that everyone gets a 157. Then that
integer is the only one that corresponds to more than one percentile group (it
corresponds to all 100 groups, in fact). However, don’t reduce further this
way. This gives exactly 1 integer, so the minimum number of integers
corresponding to more than one percentile group is 1, which is Quantity A.

As for Quantity B, though, a particular integer may have no percentile groups
corresponding to it. In the previous example, if everyone gets a 157, then no
one gets a 158, or a 200 for that matter. So the minimum number of percentile
groups corresponding to a score of 200 (or to any other particular score) is 0,
which is Quantity B.

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